Here is an example to flesh out Keerthi's nice answer in a very simple case.

Take $V$ to be the $0$-dimensional variety $V=\operatorname{Spec} F $ where $F/\mathbf{Q}$ is a finite Galois extension of degree $d$. The left hand side is given by $H^0_{dR}(V) = F$, while $V(\mathbf{C})$ is the finite set $\Sigma=\operatorname{Hom}(F,\mathbf{C})$, so that $H^0_B(V(\mathbf{C}),\mathbf{Q}) \cong \mathbf{Q}^{\Sigma}$. The comparison isomorphism is then just the usual isomorphism $\omega : F \otimes \mathbf{C} \xrightarrow{\cong} \mathbf{C}^\Sigma$.

Choosing a basis $(a_1,\ldots,a_d)$ of $F$ over $\mathbf{Q}$ and the canonical basis of $\mathbf{Q}^{\Sigma}$, the entries of the matrix of $\omega$ are just $\sigma(a_i)$ with $\sigma \in \Sigma$ and $1 \leq i \leq d$. Each time you have a polynomial relation $P(a_1,\ldots,a_d)=0$ with $P \in \mathbf{Q}[X_1,\ldots,X_d]$ (which you can interpret as an algebraic cycle on $V^d =\operatorname{Spec} F^{\otimes d}$), you get a corresponding relation on the $\sigma(a_i)$'s. As you have probably guessed, the motivic Galois group in this case is just $\operatorname{Gal}(F/\mathbf{Q})$ seen as a finite algebraic subgroup of $\mathrm{GL}(H^0_B) \cong \mathrm{GL}_{d/\mathbf{Q}}$. (NB : this "definition" of the Galois group is actually closer in spirit to Galois's original definition.)

Another example is given by an elliptic curve $E/\mathbf{Q}$ which has CM by an order in an imaginary quadratic field $K$. In this case $\mathrm{GL}(H^1_B(E)) \cong \mathrm{GL}_{2/\mathbf{Q}}$. The complex multiplication provides an algebraic cycle on $E \times E$, and the motivic Galois group is the normalizer of $K^\times$ seen as a subgroup of $\mathrm{GL}_{2/\mathbf{Q}}$.